Let $R$ be the region bounded by
$$y=1/x,\quad y=1, \quad y=2,\quad x=0$$
Consider the solid generated by rotating $R$ about the y-axis
Sketch the region, the solid, and a typical disk/washer/shell (your choice)
I know that I have to slice horizontally and integrate with respect to $y$ but I can't figure out where I am going wrong.
I got from $\displaystyle \pi\int_0^1 \frac1y dy$
Am I going about this correctly so far?
Best Answer
Hint: The volume is equal to $$\int_{y=1}^2 \pi \frac{1}{y^2}\,dy.$$
Remark: You forgot the squaring part: note that a cross-section at height $y$ has area $\pi x^2$, that is, $\frac{\pi}{y^2}$.
In your solution, there is also an issue with the limits of integration. Note that the problem says that the region is bounded in particular by the lines $y=1$ and $y=2$.
The sketch may not be correct, The region of interest is bounded above by the line $y=2$, below by the line $y=1$, on the left by $x=0$ (the $y$=axis) and on the right by $y=\frac{1}{x}$.